See Advanced Topics in elliptic curves for the full question(see also errata: http://www.math.brown.edu/~jhs/ATAEC/ATAECErrata.pdf):
2.30 (pg 184) Given $E/L$ an elliptic curve with complex multiplication by the ring of integers $R_K$ of $K$. Assume that $L$ does not contain $K$. Let $L'=LK$, so $L'$ is a quadratic extension of $L$, and let $\mathfrak{P}$ be a prime of $L$. Assume that $E$ has good reduction at $\mathfrak{P}$. Prove that $\mathfrak{P}$ is unramified in $L'$.
I think I should be using the theorem that $\mathfrak{P}$ ramifies in $L'$ $\iff$ $\mathfrak{P} \mid \text{disc}(L'/L)$, but how do I show that $\mathfrak{P}\nmid 2$? I think I would need to use good reduction, but I'm not sure how. Thanks for your help in advance!
Answering my own question:
First, I'll answer something from the comments: $L'$ is a quad extension of $L$ (see Silverman's Arithmetic of Elliptic Curves III.9.4)
Now let $\rho\in\text{Aut}(L'/L)$.
Claim 1: $K\subseteq L$ if and only if every element of $\text{End}(E)$ is defined over $L$.
$\blacktriangleleft$Let $\sigma$ be an automorphism of the algebraic closure $\overline{L}$ over $L$. Choosing $\omega$ the invariant differential defined over $L$, we have $$\omega[\alpha]^{\sigma}=\alpha^{\sigma}\omega$$ where $[\cdot]$ is defined in ATAECII.1.1 as an isomorphism such that $(E,[\cdot])$ is normalised. So $[\alpha]^{\sigma}=[\alpha]$ if and only if $\alpha^{\sigma}=\alpha$ which gives us the result. $\blacktriangleright$
Claim 2: If $\mathfrak{P}$ ramifies in $L'$, then the effect of $\rho$ is trivial on the the residue class field.
$\blacktriangleleft$ See Prove automorphism is trivial $\blacktriangleright$
Claim 3: $\text{End}(E)\rightarrow\text{End}(\tilde{E})$ is injective.
$\blacktriangleleft$ See Silverman's AECVII.2.1 $\blacktriangleright$
Now we know from the claim 1 that there exist an endomorphism $f\in\text{End}(E)$ defined over $L'$ but not $L$. So we have $f^{\rho}\neq f$. If $\mathfrak{P}$ ramifies in $L'$, then the effect of $\rho$ is trivial on the the residue class field (claim 2), so reducing mod $\mathfrak{P}$ gives $\widetilde{f^{\rho}}=\widetilde{f}$. But this contradicts with the injectivity of the reduction map on $\text{End}(E)$ (claim 3). Hence $\mathfrak{P}$ is unramified.